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Web only steel beam?

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MHSpurs

Structural
Apr 11, 2005
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Hi

Has anyone had any experience designing flat plate as a beam?

I need to design a tapered flat plate (250mm x12mm wide tapering to 150x 12mm wide) cantilever beam supporting a glass canopy.

The beam is lightly loaded but has no lateral restraint.

I cant seem to find any design guides or codes that cover the topic.

Any ideas?

 
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When I have this, I check AISC's equations for bending on a beam with top and bottom cope. I also check buckling on the plate per Roark's Stress and Strain book.
 
Hate to answer a question with a question, but since it's a cantilever, doesn't the typical analysis get thrown out the window? Cantilever wide flanges get designed differently because the normal analysis is not always give you the required reliability factor. I had to design a wideflange crane cantilever one time and did quite a bit of research on it. I would go on to assume that a cantilever plate in particular wouldnt conform to normal analysis since it has literally no lateral stiffness. At least a wide flange and a WT has a little. Maybe someone has experimental data?

 
The AISC 360-05 has a setin for rectangular plates bent abou the major axis. It's the same thing we're used to: checking yielding and LTB. You may be required to have some lateral restraint at the end or point of application of the load; you'll have to check. You could always run a FEA on it too.
 
UcfSE-
AISC 360-05 gives info in F11 on 16.1-60. This is in regards to bars - I know there is a difference between plates and bars in terms of widths and thicknesses when they are fabricated. In your opinion, can you use this section for plates since the equations really have no knowledge of what the shape was made from (whether is was from bar stock or from a plate cut down to the width of a bar)?
 
The challenge you are facing is stability - local buckling of slender flexural members. Check AISC Sections B & F and associated appendic B5 & F7, if applicable, (9th ed.) for more information.
 
kslee1000-
I gave you a star because I thought that was a valuable reply. I do have a couple comments for you, however, that I would appreciate your input on.
In the 9th edition manual, on pg 5-35, an unstiffened element is described as one which is supported on only one edge. I would think that this plate is not supported on any edge and that the width-thickness ratio limits may be unconservative as a result.
In addition, Appendix B5 doesn't list plates as an option for calculating Qs. The closest would be stems of tees, but again, I don't think this situation is addressed directly.
It does seem odd that this isn't discussed in greater detail since a great deal of stair stringers are plates.
 
Hi MHSpurs

I know of a technique which may help you however it depends on the length of the beam?
I would design it on a strain energy method like a flat leaf spring.
What length of beam have you got?

regards

desertfox
 
desertfox-
I believe that if the 250 mm leg was horizontal we would not be having this discussion. In that case, it would be a simple allowable bending stress of 0.75Fy for the 9th edition and a check for yielding in LRFD. There would be no need to concern yourself with LTB.
 
Thanks for the reply.

the 250-150 is on the vertical.

I have decided to use the basic section properties of Zx but reducing the allowable steel stress due to slenderness.

I have also talked the architect into a cable tie at the end of the cantilever and a partial torsional restraint at the internal support.

This allows the Le of the member to be reduced to within code tolerances.

Bending about x-x is not a problem.

The canopy is external so it lools like lateral loading from wind is going to govern the design.

I'm also getting someone to run an FE on it to be sure!

Still, an interesting problem, and one that warrents addressing from the codes. It seems to be a common detail for caniopies/balconies.
 
MHSpurs-
I would be careful about using Zx. Zx implies that you can develop the full plastic moment of the beam before buckling occurs.
 
MHSpurs

The only formular that I have seen for steel plates bent about their strong axis is:

Fb=.6 Fy for Ld/t^2 less that 500

Fb= 10,000/(Ld/t^2) ksi for Ld/t^2 over 500

where L is the span inches
d is the depth inches
t is thickness inches

This formular is stated to be for a simply supported beams only.

You are on your own to determine the unsupported length.
 
If using ASD, I think the b/t ratio will control. Fb = 0.6*Fy*Q. Q is based on the b/t ratio, with b equal to 1/2 the depth of the plate, because only 1/2 the depth of the plate is in compression.

DaveAtkins
 
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